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Q16E

Expert-verified
Found in: Page 434

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Determine whether the sequence converges or diverges. If it converges, find the limit.$${a_n} = \sqrt {\frac{{n + 1}}{{9n + 1}}}$$

The sequence converges and the limit is $$\frac{1}{3}$$.

See the step by step solution

Definition

A sequence$$\left\{ {{a_n}} \right\}$$has the limit $$L$$and we write $$\mathop {\lim }\limits_{n \to \infty } {a_n} = L\;\;or\;\;{a_n} \to L$$as$$n \to \infty$$if we can make the terms$${a_n}$$as close to$$L$$as we like by taking$$n$$sufficiently large. If $$\mathop {\lim }\limits_{n \to \infty } {a_n}$$exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent).

Simplify

Consider the sequence$${a_n} = \sqrt {\frac{{n + 1}}{{9n + 1}}}$$

Simplifying we have:

\begin{aligned}{a_n} &= \sqrt {\frac{{n + 1}}{{9n + 1}}} \\ &= \sqrt {\frac{{\frac{{n + 1}}{n}}}{{\frac{{9n + 1}}{n}}}} \\ &= \sqrt {\frac{{1 + \frac{1}{n}}}{{9 + \frac{1}{n}}}} \end{aligned}

Apply limit

Now evaluate the limit as$$n \to \infty$$.

\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \sqrt {\frac{{1 + \frac{1}{n}}}{{9 + \frac{1}{n}}}} &= \sqrt {\frac{1}{9}} \\ &= \frac{1}{3}\end{aligned}

Which is finite.

Hence the sequence $${a_n} = \sqrt {\frac{{n + 1}}{{9n + 1}}}$$converges and value of limit is $$\frac{1}{3}$$.